The volume of 0.0168 mol of O_(2) obtained by decomposition of KClO_(3) and collected by displacement of water is 428 ml at a pressure of 754 mm Hg at 25^(@)C. The pressure of water vapour at 25^(@)C is

The volume of 0.0168 mol of O_(2) obtained by decomposition of KClO_(3

1.	The volume of 0.0168 mol of O_(2) obtained by decomposition of KClO_(3) and collected by displacement of water is 428 ml at a pressure of 754 mm Hg at 25^(@)C. The pressure of water vapour at 25^(@)C is
Answer» 18 mm HG 20 mm Hg 22 mm Hg 245 mm Hg. Solution :Pressure of dry gas, P=nRT/V `=0.0168xx(0.0821xx1000xx760" ML "mm)xx(298)/(428)` 730 mm Pressure of moist gas=754 mm. Hence, pressure of water vapour =(754-730)mm=24 mm. Alternatively, volume of `0.0168" mol"` of `O_(2)` at STP `=0.0168xx22400=376.3" ml"` THUS, `V_(1)=376.3" ml", P_(1)=760" mm", T_(1)=273" K"` `V_(2)=428" ml",P_(2)=?,T_(2)=298" K"`. Calculate `P_(2)`.